How to add (or subtract) two exponential components with same base? I am to solve for $x$ using logs: $e^{2x}-e^x-110=0$
During my steps, I'm unsure how to combine $e^{2x}-e^x$ into one. My attempt:
$$e^{2x}-e^x-110=0$$
$$e^{2x}-e^x=110$$
Here's where I get confused:
$$\ln(e^{2x}-e^x)=\ln(110)$$
$$2x-x=\ln(110)$$
$$x=\ln(110)$$
My textbook says the solution is $\ln11$.
Where did I go wrong and how can I arrive at $\ln11$?
More specifically, how can I simplify this line on the left of the equals sign? $e^{2x}-e^x=110$
 A: Hint: let $y=e^x$, and solve a quadratic first.
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On where you went wrong, note $\log(a \pm b) = \log(a) \pm \log(b)$ does not hold true.
A: Your error starts here:
$$\ln (e^{2x}-e^x)=2x-x$$
This can be at least logarithmic equation, but not a logarithmic equality.
The correct rule is as follows:
$$\ln (e^{2x}\times e^{x})=\ln e^{2x}+\ln e^x=2x+x$$
Because, this is not a logarithmic equality:
$$\ln (a+b)=\ln a+\ln b$$
But remember that,
$$\ln(ab)=\ln a+\ln b$$
is an obvious logarithmic equality.

For the correct solution method, you can use the substitution $e ^ x = t$.
You will get,
$$t^2-t-110=0.$$
A: Brother, I know you're smart enough that you know quadratic equations,
Here in your question it is asked that :
\exp(2x)-exp(x)=110
You can subsitute \exp(x)=u
And by the knowledge of exponentials we know
\exp(2x)= u^{2}
which gives us :
\u^{2}-u=110
then you can solve this equations easily then when you get to value of u, break u into \exp(x) and then you can take the logarithm and get the value of x.
